Re: Another PHP/MySQL question

2001-06-10 Thread MikeBlezien
On Sun, 10 Jun 2001 19:13:01 -0700, "Vladimir Kravtsov" <[EMAIL PROTECTED]> wrote: >> >> >> >>>connectdb(); >>mysql_select_db(xtopsites); >>$validated=1; >>$sql = "select id, siteurl, sitename, from sitesats where validated = "$validated""; >//

Re: Another PHP/MySQL question

2001-06-10 Thread Whit Blauvelt
On Sun, Jun 10, 2001 at 07:13:01PM -0700, Vladimir Kravtsov wrote: > $sql = "select id, siteurl, sitename, from sitesats where validated = "$validated""; >// Line 8 Actually, using '$validated' might work better - one or the other. Whit -

Re: Another PHP/MySQL question

2001-06-10 Thread Whit Blauvelt
On Sun, Jun 10, 2001 at 07:13:01PM -0700, Vladimir Kravtsov wrote: > $sql = "select id, siteurl, sitename, from sitesats where validated = "$validated""; >// Line 8 > What's wrong here? Try \"$validated\" - you need to escape the quoted quotes. Whit ---

RE: Another PHP/MySQL question

2001-06-10 Thread Don Read
On 11-Jun-01 Vladimir Kravtsov wrote: > > > > connectdb(); > mysql_select_db(xtopsites); > $validated=1; > $sql = "select id, siteurl, sitename, from sitesats where validated = > "$validated""; // Line 8 ^ ^ - escape your quotes, \"$validated\"; change to sin

Re: Another PHP/MySQL question

2001-06-10 Thread Augusto Cesar Castoldi
You have a problem with "... " is reserved word, put \ before it... try this: $sql = "select id, siteurl, sitename, from sitesats where validated = \"$validated\""; regards, Augusto On Sun, 10 Jun 2001, Vladimir Kravtsov wrote: > > > > connectdb(); > mysql_select_db(xtopsites); > $valid